ATMO/OPTI 551a Gravity
Fall 2008
Gravity, Geopotential, Geoid and Satellite Orbits
Much of this introductory material on gravity comes from Turcotte and Schubert (1982).
The satellite orbit material comes from Elachi and van Zyl (2006).
Gravitational acceleration
The gravitational force between two point masses, M and m, pulling on one another is:
Fg  
GMm
rˆ
r2
where G is the gravitational constant = 6.67300 × 10-11 m3 kg-1 s-2 and r is the distance
between the centers of the two masses in meters. The minus sign means the force of
 one another. Since F = ma, the gravitational acceleration felt
gravity pulls the masses toward
by mass, m, is simply
a = GM/r2
(1)
Consider Earth’s surface gravity. We assume at first that Earth is a perfect sphere so that it can
be treated as a point mass. Given Earth’s mass =5.9736e24 kg and its average radius of
6.37101x106 m. The resulting surface gravity is 9.820660317 m/s2. This is quite close to the
standard average surface gravity of 9.80665 m/s2.
Note: GM is known far better than G or M.
Issues related to Earth Spin…
Earth is not a sphere
The Earth gravitational acceleration we feel includes the effects of the Earth’s spin. There
are two first order effects to be considered, the Earth is no longer a sphere and cannot be rtreated
as a point mass and there is centrifugal force to be considered. Because of the spin of the Earth,
the Earth has an equatorial bulge such that its equatorial radius is larger than its polar radius by
about 20 km. This means a location at the pole is actually slightly closer to the Earth’s center
than a location on the equator. Therefore we can anticipate that the surface gravity at the pole is
slightly larger than the surface gravity at the equator (see Table below).
The acceleration at the surface of Earth due to the force of gravity (which includes the
effect of the greater radius at the equator) is
g
GM 3GMa2 J 2

3sin 2  1
r2
2r 4
where a is Earth’s equatorial radius and  is the latitude and J2 is the second spherical harmonic
of Earth’s gravitational field (=1.08263e-3) which is a measure of Earth’s equatorial bulge.
Notice in this 
simple treatment there is no longitudinal dependence. In a more complex
treatment of the gravity there is longitudinal dependence as well.
Centrifugal force
From the simple harmonic oscillator, we know the centrifugal force felt at the surface of
the Earth due to the rotation of the Earth is
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g   2 s
where  is the angular velocity (=2  /86164 seconds = 7.29e-5 radians/sec) and s is the
perpendicular distance from the spin axis which is given as

s = r cos 
where r is the distance from the center of the Earth to the surface location being considered and 
is again the latitude. Therefore the centrifugal acceleration at Earth’s surface is
g   2 r cos 
The magnitude of this centrifugal acceleration is largest at the equator where it is equal to 0.0339
m/s2. This is about 0.35% of the acceleration due to the force of gravity. The gravitational
acceleration we feel at the surface
 defines the apparent local radial direction. So the component
that is relevant to gravity is the radial component which is
gr ' g cos    2 r cos2 
Note that this is positive because it is an outward or upward acceleration so it decreases the
acceleration due to the force of gravity.
The full equation of
gravity that includes the first order effect of spin is
GM 3GMa 2 J 2
g 2 
3sin 2  1  2 r cos 2 

4
r
2r
I have compared this equation to a far more sophisticated equation of gravity derived from many
years of satellite data and found it to be good to several parts in 104.
The table
below shows the contributions of the three terms.
radius
latitude
equatorial
6378.139
0
polar
6356.7523
1.570796
Units
km
radians
GM/r^2
3GMa2J2/2r4
centrifugal
-9.798280
-0.015912
0.033916
-9.864322
0.032038
0
m/s2
m/s2
m/s2
Total gravity
-9.780277
-9.832284
m/s2
So the equatorial gravity is indeed less than the polar gravity but not as much as the GM/r2 term
would imply by itself.
Sidereal day versus solar day (what’s my frame of reference)
In order to calculate the spin rate of the Earth for the centrifugal term, we need to know
the length of a day. What is the length of a day? This may seem like an obvious question but
actually it is not and the answer depends on the frame of reference you choose.
In what we normally call a year, the Earth rotates around the sun once per year. This
number of days is actually one rotation less in comparison to the number of rotations relative to
absolute space. A “sidereal” day is slightly shorter than a “solar day” because the Earth does not
have to rotate quite as far to make a full revolution relative to absolute space as it does for the
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sun to come directly overhead again. So there are 366.25 sidereal days per year compared to
365.25 solar days per year. So the length of a sidereal day is 365.25/366.25 * 86,400 sec =
86,164 sec. (Note there is a slight mistake in Elachi and Van Zyl on page 528).
Points to a location in
absolute space
Points to the sun
Change in gravity with altitude
We can examine the radial dependence of gravity by performing a Taylor expansion of
the gravitational acceleration around the value at the surface.
g
GM 3GMa 2 J 2

3sin 2  1  2 r cos 2 

2
4
r
2r
The vertical gradient of g is

g
GM 6GMa 2 J 2
 2 3 
3sin 2  1  2 cos2 

5
r
r
r
The dominant term is the first term. Consider the fractional change in g with height.

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GM 6GMa 2 J 2

3sin 2  1  2 cos 2 

3
5
g  ln g
r
r


GM
gr
r
r2
2 6a 2 J 2
 2 cos2 
2
   3 3sin  1
GM
r
r
r2
2
r is approximately 6,371,000 m.
Point mass term
dlng/dr
-3e-7

J2 term
~3e-9
Centrifugal term
-5e-10
Units
m-1
So at 10 km =104 m altitude the fractional change in g relative to the surface is 0.3%. For many
applications this can be ignored. For some high precision applications, it cannot.
Geopotential and the geoid
The potential energy of an object in Earth’s gravity field can be determined by integrating
the work done by the gravitational force in taking the object from an infinite distance to a finite
distance r from Earth
r'
W

m
 dW

m

r'
 GM 3GMa 2 J 2

2

3sin

1


dr
r2
2r 4


r'
 gdr   

W
GM GMa2 J 2


3sin 2  1 V r

3
m
r
2r

The centrifugal term has not been included in the integral because it assumes a rigid rotation with
the spin of the Earth which, at a distance of infinity, produces an infinite and unphysical
rotational velocity.
 V is known as the gravitational potential which is the gravitational potential
energy of a mass divided by its mass.
A gravity potential, U, that accounts for both gravitation and rotation we can take the
integral of the gravity equation:
GM GMa 2 J 2
1
U r  

3sin 2  1  2 r 2 cos 2 

3
r
2r
2
The “equipotential” surface and why we care
As you
move along a surface on which the gravitational potential energy changes as you
move along it, you will feel a force either pushing against you or accelerating you. Unlike a
solid surface whose strength can oppose this force (so long as the force is not stronger than the
solid’s strength), a fluid surface will respond to this force by moving and adjusting until its
surface shape changes and there is no remaining horizontal force. Therefore (in the absence of
other forces) the ocean surface is a surface along which the gravitational potential energy is
constant. Such a surface is a called an equipotential surface. The reference equipotential surface
that defines sea level is called the geoid.
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Fall 2008
The equatorial sea level geopotential (where ) is
U r  a  
GM GMJ2 1 2 2

  a  U0
a
2a
2
The polar sea level geopotential (where /2) is
GM GMa2 J2
U r  rpol  

 U0
3
rpol
rpol

The flattening or ellipticity of this geoid is defined by
f 

a  rpol
a
 1
rpol
a
rpol  a1 f 
We assume that the Earth’s sea level surface is a geopotential and set the equatorial and polar
geopotentials equal to get 

GM GMJ2 1 2 2
GM GMa2J2

  a 

3
a
2a
2
rpol
rpol
J2 1  2 a3
a a3 J2
a  a2 J2 
1 

 3 
1 2 

2 2 GM rpol rpol
rpol 
 rpol 

Subbing rpol  a1 f  and only using first order terms in f and J2 which are both very small
yields

J 1  2a3
a 
a 2 J 2 
1 
J 2 




1 2 

1

1
2 
2 

2

2 2 GM a1 f  
 a 1 f   1 f   1 f  

J2 1  2a3
J 2 
1 
 1 f 
1
 1 f  J 2
 1 f 2 
2 2 GM
 
 

f 
3J 2 1  2 a 3

2
2 GM

Plugging in values of J2 = 1.08270x10-3 and a32/GM = 3.46775x10-3 yields a value of f of
3.3579x10-3. The true value, 3.35282x10-3, is quite close.

The shape of the model geoid is nearly that of a spherical surface. Defining r0 as the
distance to the geoid surface
r0  a1 
where << 1. Subbing this into
GM GMa 2 J 2
1
U r    
3sin 2  1  2 r 2 cos 2 

3
r
2r
2

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Fall 2008
yields
U0  
GM
GMa2 J2
1
2
 3
3sin 2  1  2 a 2 1  cos 2 
3 
a1  2a 1 
2
We then set this equal to the value of Uo that we got at the equator
U0  U r  a  

GM GMJ2 1 2 2

  a
a
2a
2
We then solve for .

GM GMJ2 1 2 2
    a
1
J2
1  2a2
2
2
a
2a
2


1  cos2 
3 3sin  1
GM
a1  2a1 
2 GM
1 J
1  2a2
1
J2
1  2a2
2
2
  2 


3sin

1

1  cos2 

3 
a 2a 2 GM
a1  2a1 
2 GM

1 J2 1  2 a2
1
J2
1  2 a2
2
  


3sin  1
1 2cos2 

a 2a 2 GM
a1  2a1 3
2 GM

1

J 1 3
J2 1  2a3
1  2a3

 1   2
3sin 2  1
1 2cos2 

2 2 GM
2
2 GM
0  1 1  

0   


J 2 J 2 1 3
1  2a3
1  2a3
2
2

3sin

1

1
2

cos





 2 GM
2
2
2 GM
J2
1  2a3
1 1 33sin 2  1 
1 1 2cos2  

2
2 GM


3J 2
1  2a3
J2
1  2a3
2
2
2
 
3sin  1 2
cos   1 3sin  1
1 cos 2  


2
2 GM
2
2 GM

 3J
 J
 2a3
1  2a3
1 2 3sin 2  1
cos2  2 1 3sin 2  1
1 cos 2  

2
GM
2
2
GM


J2
1  2a3
2
3sin


1 cos2  



2 GM
 2
 3J2

 2a3
3sin 2  1
cos 2 

1
GM
 2




J2
1  2a3
J2
1  2a3
2
2
2
3sin


1
cos


3sin



 2 GM 
 2
 2 GM sin 2  
2

3
1  2 a 3  2
   J2 
sin 
2 GM 
2
So the surface of the geoid is defined as

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 3
1 a3 2  2 
r0  a1  J2 
sin 
2 GM 
 2

which is also
r0  a1 f sin 2  

A better model is
 2 f  f 2 

2

r0  a1
sin

2
1
f






1/ 2

with a = 6378.139 km and f = 1/298.256.
The actual geoid is known to much higher precision and is being updated most recently
by the Gravity RecoveryAnd Climate Experiment (GRACE). Note that GRACE actually
measures the time-evolving gravity field and geoid.
See http://earthobservatory.nasa.gov/Library/GRACE_Revised/page3.html
The geoid is critical as a reference surface that determines what the surface of the ocean should
be in the absence of any dynamics, that is ocean currents. One way currents reveal themselves
to satellites is as topography on the ocean surface relative to the geoid. This is the result of
geostrophic balance of the pressure gradient and the coriolis forces.
Satellite Orbits
For observing weather and climate related phenomena, there are 2 main classes of satellite
orbits, geosynchronous (or geostationary) and low Earth orbit (LEO).
We will discuss circular orbits primarily. While it takes 6 orbital “elements” to uniquely
define a satellite orbit, from our perspective, the main features of an orbit are its radius or
altitude, orbital period, orbital velocity, inclination, times of day crossing the equator (“longitude
of the ascending node”) and precession rate. These are not independent of one another.
Inclination
Inclination refers to the tilt of the orbital plane relative to the equatorial plane of the Earth. A
o
0 inclination has a plane coincident with the plane of Earth’s equator and orbits with the same
rotation as the Earth’s spin. A 90o inclination is exactly a polar orbit, orthogonal to the Earth’s
equatorial orbit. A 180o inclination is also coincident with Earth’s equatorial plane but the
satellite orbits in the opposite direction as the Earth’s spin and is said to be a “retrograde” orbit.
Getting into such an orbit is expensive energetically (=bigger launch vehicle) because the launch
vehicle cannot take advantage of the momentum of the Earth’s spin (~0.5 m/sec) at launch.
Choice of inclination is often set by the desired latitudinal coverage. If you want coverage
that extends from pole to pole, you need an inclination that is nearly polar or at least a high
inclination of 70-110o. Repeat cycle is another important consideration.
Further refinement of the choice will also depend on if your instrument looks downward
(nadir-viewing) or is looking at the Earth’s limb (limb-viewing). An orbit for a nadir viewing
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instrument will have to be closer to a polar orbit depending on how far off nadir the instrument
looks as it scans. If it looks exactly nadir like the CloudSat cloud profiling radar and you want
pole to pole coverage then the orbit would have to be exactly polar.
If the instruments are limb-viewing then a 70 degree orbit may suffice depending on the
altitude of the orbit. For a given radius, what is the minimum inclination, Imin, necessary for a
limb viewing instrument to see the pole from an orbit with altitude, h? The answer is
sin(Imin) = RE/r = RE/(RE + h) = 1/(1 + h/RE)
where is RE the Earth’s radius and r is the orbital radius = RE + h.
For h=750 km, Imin = 63.5o. This is the minimum inclination to
just see the pole, so the actual inclination used will be somewhat
larger.
h
RE
Orbital plane,
edge-on
I
Equatorial
plane, edge-on
Geostationary orbits
Geostationary orbits have a 24 hour period such that the satellite sits over the same longitude
of the Earth and can see the entire diurnal cycle, particularly cloud tops. Circular geostationary
orbits are equatorial orbits with an inclination of 0o.
LEO satellites
Low Earth orbits have altitudes typically ranging from 500 to 850 km and orbital periods are
about 100 minutes. A special class of near-polar orbits is sun-synchronous so they always
measure the same 2 times of day or more generally precessing where the local time of the
observations drifts. (near-)polar orbiters are used to observe high latitudes.
The lower altitude limit of usable orbits is set by atmospheric drag which causes the orbital
altitude to decay such that the orbit is not stable and the satellite will eventually reenter the
Earth’s atmosphere. The density of the atmosphere at these altitudes varies with the 11 year
solar cycle. The reason is as the sun emits more UV and higher energy radiation during the
solar cycle maximum (due to a 22 year convective-magnetic field cycle on the sun), this energy
is absorbed by the upper atmosphere causing the atmosphere to warm and expand. This pushes
the atmospheric density structure to higher altitudes at the peak of the solar cycle causing more
drag on spacecraft. The lowest altitude orbits are chosen to be closer to the surface, often for
purposes of imaging resolution, to achieve higher signal to noise ratio (SNR) for active
instruments with limited transmit power or increased sensitivity to variations in the Earth’s
gravity field. Altitudes higher than ~1000 km are avoided when possible to avoid higher
radiation levels there from energetic particles which drive up the cost of the instrument and
satellite electronics.
Orbital period for circular orbits: Harmonic oscillator
The simple motion of a satellite in a circular orbit around spherical planet can be described in
terms of sine and cosines.
y(t) = r sin(+2t/T),
x(t) = r cos(+2t/T)
where T is the orbital period and r is the orbital radius and t is time. The + refers to the direction
the satellite moves in its orbit, clockwise or counterclockwise. For the moment we are interested
in magnitudes so we’ll drop the +. The velocities components are
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vy(t) = 2 r/T cos(2t/T),
vx(t) = - 2 r/T sin(2t/T)
(2)
ax(t) = - r (2/T)2 cos(2t/T)
(3)
The centripetal accelerations are
ay(t) = -r (2/T)2 sin(2t/T),
We set the magnitude of this centripetal acceleration, r (2/T)2, equal to the gravitational
acceleration to determine the orbital period versus radius
r (2/T)2 = GM/r2
from which we get Kepler’s famous law: r3 T2:
T2 = (2)2 r3/GM
Alternatively we can write the gravitational acceleration in terms of the surface gravitational
acceleration, gs, which on average is 9.81 m/sec2 for the Earth.
a = GM/r2 = gs RE2/r2
where RE is the radius of the planet in general or in this case the Earth. Again for a circular orbit,
the radial gravitational acceleration equals the magnitude of the centripetal acceleration
r3 (2/T)2 = gs RE2
In this form,
T = 2 (r3/2/RE)/gs1/2
r = [gs RE2(T/2)2]1/3
or
(4)
Orbital Period vs. Altitude
120
24
T (minutes)
115
Orbital Period (hours)
geostationary
12
6
LEO
GPS
Orbital period (minutes)
T (hours)
18
110
LEO
105
100
95
90
0
0
5000
10000
15000
20000
25000
30000
35000
40000
500
700
900
1100
1300
1500
Orbital altitude (km)
Altltude (km)
Orbital Velocity
From equation (2), the magnitude of the orbital velocity, V0, is 2 r/T. Combining this with
(4) yields
V0 = 2 r/T = 2 r gs1/2 /[2 (r3/2/RE)] = RE r-1/2 gs1/2
{Check units: m m-1/2 m1/2 s-1 = m/s }
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LEO velocity for circular orbits
7.7
ORbital velocity (km/s)
7.6
7.5
7.4
7.3
7.2
7.1
500
700
900
1100
1300
1500
Orbital altitude (km)
As the Figure shows, LEO velocities are around 7 km/sec (actually 7.5 km/sec for the typical
altitude range of 500 to 850 km altitude). The velocity scales inversely with the square root of
the orbital radius, so the velocity decreases as the orbital radius increases. The velocity of a
geosynchronous orbit is about (7/42)1/2 or 40% of that of a LEO or 3 km/sec.
We care about orbital velocity because from LEO it limits how long we can view a location
on Earth. Our instruments must be designed with this in mind.
Orbital Precession
The orientation of the orbital plane is defined relative to absolute space (NOT relative to the
changing direction from the center of the Earth to the center of the sun). The orbital plane
precesses in general depending primarily on the orbital inclination and the J2 of the planet which
is the second zonal harmonic of the Earth’s geopotential field. J2 is essentially a measure of the
equatorial bulge. The precession rate is given as
d/dt = -3/2 J2 RE3 gs1/2 cos(I)/r7/2
(6)
where is the longitude of the ascending node, and I is the orbital inclination
check units: m3 m1/2 s-1 m-7/2 = s-1 = rad/sec
Additional points:
 Note that a perfectly polar orbit (I = 90o) does not precess.
 For an orbit to precess in the same direction as Earth’s spin, its inclination, I, must be
larger than 90o.
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Sun synchronous orbits:
A sun synchronous orbit keeps the alignment of the orbital plane fixed relative to the line
between the center of the Earth and the center of the Sun. Sun synchronous orbits are desirable
for keeping the solar illumination the same from orbit to orbit which simplifies satellite and
instrument designs.
WEATHER:
Large, “polar-orbiting”, weather satellites carrying many instruments are
often in sun synchronous orbits. Strictly speaking they are not in polar orbits but their
inclination is close to 90o as we will see. Their orbits are often described by the time of day they
cross the equator.
CLIMATE: They used to be the orbits of choice for LEO climate measurements because the
observations don’t drift in local time of day so the diurnal cycle in theory does not enter into the
long term measurements. However, sun synchronous orbits can drift slightly over time, causing
the diurnal signal to alias into the long term climate signal causing a nightmare to try to remove
such a subtle signal while looking for another subtle signal. Another problem is the diurnal cycle
itself is predicted to change and is apparently changing as the climate warms. Such changes are
guaranteed to alias into long term trends measured by sun synchronous orbits. In my opinion,
orbiting observing systems should be designed to sample all times of day throughout the year so
that the diurnal signal and the seasonal signal are captured as part of the climate signal. Then
clever researchers can separate the different time dependent signals apart during their analysis
as they try to unravel what the climate system is actually doing.
To make an orbit sun-synchronous, the orbit must precess one extra revolution in a year. So
the precession rate is 360o in 366.25 sidereal days or about 1 degree per day or 0.2 microrad/sec.
We set eqn. (7) equal to 1.99x10-7 rad/sec to get the Figure below.
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Sun Synchronous Orbit
102
101.5
Orbital inclination (degrees)
101
100.5
100
99.5
99
98.5
98
97.5
97
500
600
700
800
900
1000
1100
1200
1300
1400
1500
Orbital altitude (km)
Figure A. Inclination vs. orbital altitude for sun synchronous orbits
Example:
CloudSat moves in a sun-synchronous orbit which has an equatorial altitude of
approximately 705 km. This sun-synchronous orbit is nearly circular and is inclined with respect
to the earth's equator at 98.2 degrees (see Figure A). The CloudSat orbit is stable for 20 to 30
years before drag will bring it down into the Earth’s atmosphere.
Sampling and coverage
Geosynchronous orbiters provide essentially continuous coverage of the region underneath
them. Polar orbiters essentially sample two times per day (one day and one night) at the equator,
as the orbit slowly precesses and the Earth rotates underneath the orbital plane. For sun
synchronous orbits, the solar times are fixed. For all other LEOs, the solar time of the two
equatorial crosses drifts with time.
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For orbital altitudes between 500 and 1500 km, the orbital period varies from about 95 to 115
minutes. Since a solar day is 1440 minutes in length, the number of daytime equatorial crossings
per day is about 13-15.
For an altitude of 550 km, the number of orbits per day is about 15. So the spacing between
daytime (or nighttime) equatorial crossings is 360o/15 = 24o of longitude. So if your nadirviewing instrument can sweep back and forth by +12 degrees of longitude, your instrument could
sample the entire globe every day. 1o of longitude at the equator is about 111 km so 12o is 1332
km. The look angle off nadir would have to be tan() = 1332/550 so  = +68o which is quite
large. A higher orbit would achieve full global coverage (at a cost of resolution because it is
higher above the Earth). Because of the wide swaths required to achieve full coverage each day,
there is typically a gap between consecutive swaths at the equator.
AIRS example
The Atmospheric InfraRed Sounder (AIRS) is a nadir-viewing, high resolution IR
spectrometer on NASA’s AQUA satellite in the A-train (see below) flying at 705 km altitude in a
sun synchronous orbit. The AIRS infrared bands have an instantaneous field of view (IFOV) of
1.1º (=19 mrad) and FOV = ± 49.5º (=0.86 rad) scanning capability perpendicular to the
spacecraft ground track. So the horizontal resolution is tan(0.019)*705 km = 13.5 km at nadir
and the swath width = 2*tan(0.086)*705 km = 1650 km.
Does this swath width allow AIRS to sample the entire equator each day? At 705 km
altitude, the orbital period is 98 minutes. In 98 minutes the Earth rotates 24.5 degrees of
longitude which means the location under AIRS has moved 24.5*111 km = 2720 km. So each
day AIRS samples the atmosphere about 1650/2720 = 60% of the equatorial area.
The combined high spatial resolution and wide swath width combined with the fast orbital
motion of 7.5 km/sec means there is not a lot of time for each measurement. The measurement
integration time for each 13.5 km sounding of the atmosphere must be quite short given the 7.5
km orbital velocity of AQUA and AIRS. The time it takes the satellite to move 13.5 km along
its orbital track is 1.8 seconds. This is the time available to do a scan across the full swath width.
So the number of individual 1.1o footprint soundings per swath width scan is 90 and the time of
data measurement per sounding is 1.8 seconds/90 = 20 msec. The actual integration time is
22.41 ms for each footprint of 1.1º in diameter. This short time likely hurts the signal to noise
ratio (SNR) a bit and reduces the accuracy of the individual profiles but AIRS is trying to do a
lot and tradeoffs must be made.
AIRS is a very powerful sounder sampling a large portion of the global atmosphere each
data. However, since AIRS is an IR instrument, it requires clear sky to derive atmospheric
profiles and about 95% of the AIRS footprints are cloud contaminated (Joanna Joiner, pers.
comm.) which reduces its actual coverage to about 5% of its theoretical coverage. In regions that
are systematically cloudy, AIRS may have trouble profiling the atmosphere. Still AIRS is a
VERY powerful atmospheric sounder for temperature, water vapor and other trace constituents
in the atmosphere.
Orbital Repeat Period
Elachi and van Zyl (EvZ)’s Figure B-7 shows repeat periods for LEO sun synchronous
orbits, that is, the time and the number of orbits between which the satellite flies exactly over the
same location again. The easiest way to understand this figure is to start with lowest row of
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numbers that has 5 items reading R=16 R=15 R=14 R=13 R=12. This row
coincides/aligns with the y-axis cycle period of 1 day. This means that at R=16, a satellite with
an orbital altitude of approximately 270 km will fly over exactly the same locations on the Earth
one day later after the satellite has gone around the Earth 16 times.
The second row has entries 31 29 27 and 25. The 29 means that for an altitude of about
710 km, the satellite will fly over the exact same locations every 2 days after orbiting the Earth
29 times.
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Semi-arid satellite mission?
This past summer we looked briefly at the feasibility and utility of a semi-arid land LEO
satellite that would sample the North American Southwest as well as other semi-aird and arid
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regions of the globe as often as possible. From EvZ’s figure B-7, the fastest repeat time for a
LEO in the 500 to 900 km range is once per day at either 560 km (15 orbits per day) or 900 km
(14 orbits per day). A basic problem is that an enormous amount of funds (several hundred
million dollars) would be required to bring such a mission to reality and only two times of day
would be sampled in a region where the diurnal cycle is very large and very important.
Geosynchronous satellite would be better in terms of coverage but far more expensive and would
have to confront the standard problem of fine horizontal resolution from 36,000 km in space.
Particularly during the monsoon, the observations that could penetrate the cloud cover would be
quite limited. We discussed stratospheric, lighter than air platforms being developed for telecom
and Star Wars applications but they still have some serious technical problems to overcome.
Repeating orbits for calibration: TOPEX and JASON
Systematically flying over the exact same location every so often can be very important for
calibration of the satellite observations. TOPEX and its successor, JASON, carry altimeters to
measure the ocean topography for inferring ocean currents and warm and cold regions for severe
weather, and large scale waves like Kelvin waves associated with the El Nino-Southern
Oscillation (ENSO) cycle and the long term rise in sea level predicted (with large error bars) to
occur with global warming. In order to make sure the ocean sea level measurements are right,
these orbits are designed to repeat every so often over locations with very precise sea level
gauges whose measurements can be compared with the altimeters to quantify and understand the
errors. This has worked quite well and satellite altimeter measurements of the sea level rise
since 1992 when TOPEX launched are very good and indicate 2.8±0.4 mm/yr. Our
understanding of what is contributing to the rise is less certain. The dominant contributor is
probably thermal expansion of the upper oceans but how much is due to melting land ice is
unclear.
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To see how the sea level has been changing by region, see also
http://globalclimatechange.jpl.nasa.gov/news/index.cfm?FuseAction=ShowNews&NewsID=16
NASA’s A-train (see: http://events.eoportal.org/pres_AquaMissionEOSPM1.html) consists
of several satellites each in the same orbit but slightly delayed with respect to one another. The
orbit is summarized as Sun-synchronous circular orbit, altitude = 705 km (nominal), inclination
= 98.2º, local equator crossing at 13:30 (1:30 PM) on ascending node, period = 98.8 minutes, the
repeat cycle is 16 days (233 orbits). The repeat period allows flights over calibrating ground
instruments every 16 days to help calibrate the orbiting instruments and refine retrieval
algorithms.
The long time between repeat flights allows the Earth’s surface at the equator to be carved up
into 233 longitude sections, the longitudinal width of which depends on each instrument. For a
passive instrument like AIRS with its wide 1650 km or 15o of longitude coverage every orbit and
60% sampling of the globe each day, the 233 sections is not terribly important. But for the 94
GHz cloud profiling radar (CPR) on CloudSat (also in the A-train) which can only look straight
down, carving the equatorial Earth into 233 longitudinally-narrow strips is quite relevant. The
cross-track resolution of the CPR is 1.2 km. So every 16 days, the CPR covers 233*1.2 km =
280 km of longitude or about 0.7% or the equatorial region (systematically never sampling the
rest of the equatorial longitudes).
The CALIOP LIDAR on CALIPSO (also in the A-train) has a 90 m instantaneous footprint
which is smeared to 333 m in the along track direction by orbital motion over the LIDAR pulse
duration. CALIOP looks straight down so there is no scanning to produce a larger swath width.
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So every 16 days, CALIOP covers 233*90 m = 21 km of longitude or about 0.05% or the
equatorial region (again, systematically never sampling the rest of the equatorial longitudes).
The power of these instruments lies not in the horizontal coverage but rather the unique
vertical information they provide. CPR provides 500 m vertical resolution and can penetrate
through clouds to give the first 3D information on clouds globally. CALIOP provides 30 m
vertial resolution profiling of aerosols and clouds, far better than the 1 to 4 km can be achieved
with passive measurements. One hopes the sampling by these two high vertical resolution
instruments, CPR and CALIOP, will produce a statistically representative sampling of the
equatorial region. These are key examples of the tradeoffs one must make with orbiting active
instruments.
References:
See Turcotte and Schubert.
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